Question

Position time equation for a particle is $x=2t^3 -6t^2$. Its maximum speed will be:
A. $6\;ms^{-1}$
B. $12\;ms^{-1}$
C. $18\;ms^{-1}$
D. $3\;ms^{-1}$

Answer 1

Hint: Differentiate the position-time equation twice to get an expression for acceleration. Equate it to zero and find the instant t at which it becomes 0. This will also be the instant where the particle possesses the maximum speed. Differentiate the position time equation once to get a velocity expression and substitute the t value in it to arrive at its maximum speed.

Formula Used:
Velocity $v(t) = \dfrac{dx}{dt}$
Acceleration $a(t) = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$

Complete Solution:
We are given with a position-time equation $x(t) = 2t^3-6t^2$

We know that the speed of a particle is the rate at which its position changes, i.e., velocity is the time derivative of position, so we have:
$v(t) = x^{\prime}(t) = \dfrac{d}{dt}\left(2t^3-6t^2\right) = 2\left(3t^2\right)-6(2t)$
$\Rightarrow v(t) = 6t^2-12t$

We also know that the acceleration of a particle is the rate at which the velocity of the particle changes with time, i.e., acceleration is the time derivative of velocity, so we have:
$a(t) = v^{\prime}(t) = \dfrac{d}{dt}(6t^2-12t) = 6(2t)-12$
$\Rightarrow a(t) = 12t-12$
Now, we are required to find the maximum speed of the particle.

We know that a particle can attain its maximum speed when it can no longer accelerate. This means that the instant at which the particle is unable to accelerate anymore is also the instant at which it attains a maximum speed.

Let us find the instant t at which $a(t)t=0\;ms^{-2}$ by plugging this into the
acceleration-time equation:
$0=12t-12 \Rightarrow 12t=12 \Rightarrow t = \dfrac{12}{12} = 1\;s$
This means that at $t=1\;s$ the particle ceases to accelerate since it has attained a maximum speed.

To find the magnitude of this speed, we plug in $t=1\;s$ in the velocity-time equation and take the absolute value of the solution (since speed is just the magnitude part of velocity).
Maximum speed $s_{max} = |v(1)| = |6(1)^2-12(1)| = |6-12|= |-6| = 6\;ms^{-1} $

Therefore, the correct choice would be A. $6\;ms^{-1}$.

Note:
Remember that the maximum speed is attained when the velocity no longer changes or when the change in velocity is unable to bring about any change in acceleration. It is also essential to remember that velocity is the time derivative of position, and acceleration is the time derivative of velocity, or a double time derivative of position.